The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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Construction of the field of real numbers within $ZF$ [duplicate]

I am interested in a problem whether the field of real numbers can be constructed within $ZF$. I will state the problem more precisely as follows. Definition 1 An ordered field $K$ is called ...
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Is the Axiom of Choice implicitly used when defining a binary operation on a quotient object?

Let's say you have a group $(G,\cdot)$ and you have a normal subgroup $N$ (note we are considering this only as a set). And now we want to define a binary operation $\star$ on $G/N$ such that $(G/N, ...
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Proof that a product of two quasi-compact spaces is quasi-compact without Axiom of Choice [on hold]

A topological space is called quasi-compact if every open cover of it has a finite subcover. Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$. The usual proof that $Z$ is quasi-compact uses a ...
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1answer
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What is wrong with the following proof saying Zorn's lemma implies Hausdorff maximum principle?

I am reading 'Topology' by J.R. Munkres's first chapter on set theory. In the exercises 5-7 on page 72 he asks the reader to show that Zorn's lemma implies Hausdorff maximum principle via the ...
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1answer
142 views

Can we prove that a bounded closed subset of $\mathbb R^n$ is compact without Axiom of Choice?

Can we prove that a bounded closed subset of $\mathbb R^n(n \ge 1)$ is compact without using Axiom of Choice? This is a related question which was closed.
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Can we prove that every real number is a limit of a sequence of rational numbers without Axiom of Choice?

Can we prove that every real number is a limit of a sequence of rational numbers without using Axiom of Choice?
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1answer
88 views

“Hidden” axiom of choice?

Let $\mu$ be a measure on $S$ such that: $\mu\left(\emptyset\right)=0$ and $\mu(S)=1$ if $X\subseteq Y$, then $\mu(X)\leq\mu(Y)$ $\mu\left(\{a\}\right)=0$ for all $a\in S$ if $X_n$, ...
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1answer
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Can all theorems of $\sf ZFC$ about the natural numbers be proven in $\sf ZF$?

I know a proof of Hindman's theorem that uses ultrafilters on the natural numbers, and ultimately, the axiom of choice. But the theorem itself is essentially a combinatorial property of the natural ...
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2answers
31 views

Does defining the closure of a set as the intersection of all closed set that contain it requires the axiom of choice?

Given a set $S$, the closure of $S$ is sometimes defined as the intersection of all the closed sets that contain it. This type of argument is pervasive in mathematics when one want to construct the ...
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2answers
59 views

An equivalence of AC

I have to prove the following: In $ZF^-$ the axiom of choice implies: For every set X there exist $Y \subseteq \bigcup X$ such that: Y has at most one element in common with each of X Y is maximal ...
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2answers
43 views

Is this a basis for the dual space?

There is an example on Wikipedia that I don't understand and I'd appreciate some help. They define $\mathbb R^\infty$ to be the space of all sequences that are zero except for finitely many indexes. ...
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Existence of a real uncountable $\aleph_{\alpha}$ without $\mathsf{AC}$

Set theory (Jech) $\text{p.}\;27:$ It is an open problem whether one can prove without the axiom of choice that there exists a regular uncountable $\aleph_{\alpha}\;($the informed guess is that ...
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1answer
57 views

Book/Books leading up to the the axiom of choice?

I am familiar with the axioms of ZF set theory and some basic uses of them to completely formally construct more complex objects such as natural numbers etc. However I have pretty much no background ...
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2answers
76 views

Confusion regarding one formulation of the Axiom of Choice.

One formulation of the Axiom of Choice is: The Cartesian product of non-empty sets is always non-empty. Cartesian product is defined as making "every possible pair" between elements of two sets. ...
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1answer
114 views

Unions and the axiom of choice.

Is the following equivalent to the axiom of choice? Let $A = \{a_i: i \in I\}$ be a collection of pairwise-disjoint non-empty sets indexed by $I$. Similarly, let $B = \{b_i : i \in I \}$. Further ...
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2answers
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Proving there is a sequence convergent to a limit point of a set without axiom of countable choice?

Often, we use a construction like this: Given a subset $ A $ of a metric space and its limit point $ a $, we know that for every $ \epsilon > 0 $ there is another point $ x $ different from $ a $ ...
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Amount to choice necessary to prove instances of Tychonoff theorem

Let $I$ be a fixed nonempty set. I would like to know how much choice is necessary in order to prove that the product of any $I$-indexed family of compact topological spaces is compact (under the ...
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96 views

surjective map and cardinality

I work in $\mathsf{ZF}$(without the axiom of choice). Let $A, B$ be sets such that $\left| A \right |$ and $\left|B \right |$ are both defined and let $f \colon A \to B$ a surjective function. Can I ...
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2answers
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Is the well-ordering in the well-ordering axiom required to be definable?

It is well-known that Axiom of Choice is equivalent to the statement that every set can be well-ordered. Now, to show that $M\models AC$, is it sufficient to show that there exists some well-order of ...
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4answers
65 views

Intuitive idea of axiom of choice

I'm currently reading a book on set theory and it gives the following formulation of the axiom of choice: Let $X$ be a non-empty set. Then there is a function $g: ...
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1answer
68 views

Does restricting the range of a collection of nonempty sets to one dominated by the index set require the Axiom of Choice?

The title was difficult to write, because it is hard to say the property I am looking for in words. Here it is in symbols: $$\forall i\in I\ A_i\ne\emptyset\implies\exists X\preceq I\ \forall i\in I\ ...
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The axiom of countable choice.

There seem to be many questions along the same line but none of them seem to be quite what I am asking, so here goes: If a set is countable, then we know that a bijection exists between it and the ...
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2answers
143 views

Is (countable) AC necessary for a useful theory of Lebesgue measure?

I am working through some notes on the Lebesgue measure, and I noticed that the proof that $\lambda^*$ (the outer measure) is countably sub-additive requires countable choice. (Short version of the ...
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How is it possible that the well-ordering theorem is strictly stronger than the axiom of choice in second-order logic? [duplicate]

If I am not wrong, the well-ordering theorem is strictly stronger than the axiom of choice in second-order logic. I am not sure to understand how this is possible. The reason is that second order ...
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Countable axiom of choice: why you can't prove it from just ZF

This is a follow-up question to the discussion about the finite axiom of choice here. Suppose we have a countable collection of non-empty sets $\{A_1, A_2, A_3,\cdots\}$ Reasoning as indicated in ...
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1answer
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Axiom of Choice and Zorn's Lemma Equivalence: some intuition

$$ \text{Axiom of Choice $\Rightarrow$ Zorn's Lemma } $$ $$\text{Axiom of Choice $\Leftarrow$ Zorn's Lemma } $$ I feel mathmatically immature to go through these proofs now. My quesiton therefore is: ...
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Why can't you pick socks using coin flips?

I'm teaching myself axiomatic set theory and I'm having some trouble getting my head around the axiom of choice. I (think I) understand what the axiom says, but I don't get why it is so 'contentious', ...
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2answers
135 views

Can we prove that a quasi-compact locally noetherian space is noetherian without Axiom of Choice?

I will state some definitions to clarify my question. Definition 1 Let $X$ be a topological space. If every open cover of $X$ has a finite subcover, $X$ is called quasi-compact. Definition 2 Let $X$ ...
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How can I explain to my professor his argument invokes the AC?

This is not the standard definition, but my topology professor restricted contexts in metric spaces. Definition An open set $U$ in a metric space $X$ is a subset of $X$ such that the interior ...
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Intuition for “the existence of a basis for every vector space is equivalent to the Axiom of Choice”?

Is there a intuitive way to understand "the existence of a basis for every vector space is equivalent to the Axiom of Choice"?
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1answer
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What is necessary for having a free ultrafilter?

Without any choice axioms, are there free ultrafilters on the natural numbers? If not, can we prove the existence of ANY free ultrafilters, on any set?
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1answer
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Do we need AC to prove Principle of Dependent Choices

For any nonempty set $X$ and any entire binary relation $R$ on $X$, there is a sequence $(x_n)$ in $X$ such that $x_nRx_{n+1}$ for each $n \in \mathbb{N}$. (Here an entire binary relation on $X$ is ...
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Assuming the axiom of choice, how to find explicit group structure of a given set

Let us assume the axiom of choice. This is equivalent to every nonempty set having group structure. My question is, given some nonempty set, can we define the binary operator in a constructive way ...
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1answer
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existence of spanning trees in complete graphs implies choice?

it is known that the existence of spanning trees in arbitrary (connected) graphs implies the Axiom of Choice. I was wondering if this result still holds if we restrict ourselves to spanning trees of ...
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1answer
47 views

Cardinality of union of pairwise disjoint elements needs choice?

If there is an indexed family $(i\mapsto A_i)_{i\in I}$ of pairwise disjoint sets $A_i$, why do we need choice to show that $$ \left|\textstyle{\bigcup_{i\in I}A_i}\right| = \sum_{i\in I}|A_i|? $$ It ...
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Axiom of choice in proof of Wigner's theorem?

In Appendix A of chapter 2 of "The Quantum Theory of Fields," vol. 1, Weinberg presents a proof of Wigner's theorem: given a symmetry transformation $T$ of rays, one can extend this to a symmetry ...
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The regularity of successor cardinal

I was looking at two different proofs of the fact that successor cardinals are regular. It struck me as odd that both proofs used AC. Looking at the concepts involved in defining cofinality I feel as ...
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Uncountable Cardinals without AC

I am doing an exercise, proving that without AC or Replacement that there are uncountable cardinals. As a point of reference I looked at the proof in Kunen's "The Foundations of Mathematics" that ...
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Set Theory , Konig's Lemma and Infinite Graph Theory

I am trying to understand the basics of Infinite Graph theory and various preconditions in Konig's Lemma. The texts I have studied tend to use the Axiom of Choice (usually Zorn's Lemma) as a tool of ...
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1answer
80 views

Infinite dimensional vector space. Linearly independent subsets and spanning subsets

This question is a follow up to this question: The dimension of the real continuous functions as a vector space over $\mathbb{R}$ is not countable? I realized that the answer I accepted used an ...
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1answer
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Noetherianess of a locally noetherian affine scheme without axiom of choice

I use the definition of a noetherian ring given by Qiaochu in this: A commutative ring is noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is ...
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Understanding Zorn's lemma.

A lot of authors assume Zorn's lemma. I am told it is not an obvious mathematical fact, but I am having problems understanding why that is. Zorn's lemma states that if every chain in a partially ...
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1answer
76 views

Axiom of choice and an example of a Well-ordered $\Bbb R$

From the axiom of choice we get that every set can be ordered in a way that will make it a well ordered set, including $\Bbb R$. However, since the ordinal of such a well-ordered set of $\Bbb R$ will ...
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1answer
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A well ordering on $\mathbb{R}$ and bigger sets

Consider the set of sequences $S = \{f:\mathbb{N}\to\mathbb{N}\}$, define an order on $S$ by the following: Based on the well-ordering of $\mathbb{N}$ and induction, either $f_1 = f_2$ or there is a ...
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Riemann Integration and the Axiom of Choice

In Riemann's definition of integration in $[a,b]$, a step in the process consists of choosing one point from each part of the "current" partition for further a partition, and again choose one point ...
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Zorn's lemma problem

Let $n$ be a positive natural number. Prove using Zorn's lemma that there is a set A of points in the plane that satisfies: 1. Any line in the plane does not contain $n+1$ points of A. 2. For every ...
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On Counted Languages

In my recent question on Godel Completeness I mentioned that there was a related question I wanted to ask, but would keep separate. I have been recently studying "non-well ordered sets" and Chapter 7 ...
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Constructiveness of Proof of Godel's Completeness Theorem

As a mathematician interested in novel applications I am trying to gain a deeper understanding of (the non-constructiveness of) Godel's Completeness Theorem and have recently studying two texts: ...
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When do surjections split in ZF? Two surjections imply bijection?

We have that the Axiom of Choice is equivalent to the principle that every surjection has a right inverse. However, without the Axiom of Choice we can determine for some $X$ that $X\succeq Y\implies ...
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Which sets are well-orderable without Axiom of Choice?

I know that, assuming Axiom of Choice, every set is well-orderable. I know also that the assertion that $\mathbb{R}$ is NOT well-orderable is consistent with ZF. How can I find other sets such that, ...